Foundation type selection
Check for Isolated footing
Column size= (400*400)mm2
Total number of column=16
Service load,P=1508.51
Design Load,Pu=2262.77KN
Safe bearing capacity of soil(SBC)=150KN/m2
Now,
𝑆𝑒𝑟𝑣𝑖𝑐𝑒 𝑙𝑜𝑎𝑑
Approximate area=1.10*
𝑆𝐵𝐶
A=1.1*1508.51/150
=11.062 m2
Let us consider a square footing. Then,
Size of footing= 3.326m
Therefore providing 3.5*3.5 m2 square footing.
𝐹𝑎𝑐𝑡𝑜𝑟𝑒𝑑 𝑙𝑜𝑎𝑑
Net pressure on soil=𝐴𝑟𝑒𝑎 𝑜𝑓 𝑓𝑜𝑜𝑡𝑖𝑛𝑔
=123.14KN/m2 <SBC
Therefore area occupied by 16 footing= 16*3.5*3.5
=196m2
Length=14.262m
Breadth= 14.5668m
Plinth area of block=207.75m2
50% of area block=103.875m2
Since area occupied by isolated footing is greater than 50% of plinth area,
So, Mat foundation is necessary
Determination of eccentricity
For locating geometric centroid
Column Area About column 1-c A*X A*Y
X Y
1C 0.16 0 0 0 0
1H 0.16 5.461 0 0.87376 0
1K 0.16 9.5758 0 1.532128 0
1M 0.16 13.462 0 2.15392 0
2C 0.16 0 4.7244 0 0.755904
2H 0.16 5.461 4.7244 0.87376 0.755904
2K 0.16 9.5758 4.7244 1.532128 0.755904
2M 0.16 13.462 4.7244 2.15392 0.755904
6C 0.16 0 9.0932 0 1.454912
6H 0.16 5.461 9.0932 0.87376 1.454912
6K 0.16 9.5758 9.0932 1.532128 1.454912
6M 0.16 13.462 9.0932 2.15392 1.454912
7C 0.16 0 13.7668 0 2.202688
7H 0.16 5.461 13.7668 0.87376 2.202688
7K 0.16 9.5758 13.7668 1.532128 2.202688
7M 0.16 13.462 13.7668 2.15392 2.202688
Total 2.56 18.23923 17.65402
ΣAi Xi
x= =7.1247m
ΣAi
ΣAi Yi
y= =6.8961m
ΣAi
For locating centroid of resultant forces
` Combination Axial load About Column 1-c P*X P*y
KN X,m Y,m
1C 1.5(DL+LL) 668.74 0 0 0 0
1H 1.5(DL+LL) 723.104 5.461 0 3948.870944 0
1K 1.5(DL+LL) 698.563 9.5758 0 6689.299575 0
1M 1.5(DL+LL) 572.894 13.462 0 7712.299028 0
2C 1.5(DL+LL) 1419.876 0 4.7244 0 6708.062174
2H 1.5(DL+LL) 1097.176 5.461 4.7244 5991.678136 5183.498294
2K 1.5(DL+LL) 1809.752 9.5758 4.7244 17329.8232 8549.992349
2M 1.5(DL+LL) 1891.618 13.462 4.7244 25464.96152 8936.760079
6C 1.5(DL+LL) 1479.808 0 9.0932 0 13456.19011
6H 1.5(DL+LL) 1172.825 5.461 9.0932 6404.797325 10664.73229
6K 1.5(DL+LL) 1932.998 9.5758 9.0932 18510.00225 17577.13741
6M 1.5(DL+LL) 1906.873 13.462 9.0932 25670.32433 17339.57756
7C 1.5(DL+LL) 1781.029 0 13.7668 0 24519.07004
7H 1.5(DL+LL) 2263.77 5.461 13.7668 12362.44797 31164.86884
7K 1.5(DL+LL) 1881.419 9.5758 13.7668 18016.09206 25901.11909
7M 1.5(DL+LL) 1330.914 13.462 13.7668 17916.76427 18322.42686
Total 22631.359 166017.3606 188323.4351
C.G from load
ΣP X
X = i i =7.335722m
̅
ΣPi
̅ = ΣPi Yi =8.321349m
Y
ΣPi
Eccentricity
̅ -x = 0.211022m
ex = X
̅ – y =1.425249m
ey = Y
Moment of inertia
bd3
Ix = + Ahy2
12
14.262∗14.56683
= + 196.380*(7.0834-7.3357)2
12
=3686.099 m4
db3 14.5668∗14.2623
Iy = + Ahx2 = + 196.380*(6.931-8.3213)2 =3901.061 m4
12 12
Calculation of stress
Safe bearing capacity of soil= 150KN/m2
Stress,s=(P/A)±(MY/IY)*X±(MX/IX)*Y
Area(A)=14.262*14.5688
=207.78m2
P/A=108.919696KN/m2
MX=P*ey
=23466.45615KNm
My=P*ex
=4632.639187KNm
Co-ordinate of corner of Mat Foundation and corresponding corner stress
ID About C.G of Mat Foundation Stress
X,m Y,m REMARKS
Corner1 7.131 7.284 163.7831>SBC
Corner2 7.131 -7.284 72.048<SBC
Corner3 -7.131 -7.284 54.056<SBC
Corner4 -7.131 7.284 145.7911<SBC
1C -6.731 -6.884 <SBC
57.08
1H -1.27 -6.884 63.969<SBC
1K 2.8448 -6.884 <SBC
69.16
1M 6.731 -6.884 <SBC
74.06
2C -6.731 -2.1596 86.8293<SBC
2H -1.27 -2.1596 93.718<SBC
2K 2.8448 -2.1596 98.909<SBC
2M 6.731 -2.1596 103.81<SBC
6C -6.731 2.2092 114.339<SBC
6H -1.27 2.2092 121.229<SBC
6K 2.8448 2.2092 126.42<SBC
6M 6.731 2.2092 131.322<SBC
7C -6.731 6.8828 143.768<SBC
7H -1.27 6.8828 150.65817 >SBC
7K 2.8448 6.8828 155.85 >SBC
7M 6.731 6.8828 160.75 >SBC
Design of Mat Foundation
Reference S.N Calculation Output
Strip 7-7 1 Known Data q=163.7831KN/m2
Average upward soil pressure,q=163.7831KN/m2
Maxium span length,L=5.461
IS 456-2000 2 Moment calculation Ms=488.442
Table 12 Maxium support moment , Ms=qL2/10
Ms=488.442KNm
IS 456-2000 Maxium span moment, Msp= qL2/12 Msp=407.035
Table 12 Msp=407.035MNm
SP 16 3 Depth from Monent consideration
TABLE D Depth of footing,
2 𝑀 2 488.442∗10^6
d= √ =√ =382.987mm
3.33∗𝑏 3.33∗1000
4 Permissible Punching shear
Tv’=β*.25*√fck
β =1
Tv’= 1*.25*√25
=1.25 N/mm2
IS 456-2000 5 Check for two way shear i.e. Punching shear
For corner column C1
Perimeter, b0=(d+1200) mm
Pu =2263.77KN
Nominal Shear stress
Pu
Tv=
b0∗d
Therefore, d=873.309 mm d=873.309 mm
6 So taking depth of the mat foundation, d=900mm
d=900mm
Providing effective cover 60mm
D= (900+60)mm D=960mm
=960mm
IS 456-2000 7 Calculation of Area of steel
CL. 26.5.2 Min Ast= .12% 0f bD
=.12/100* 1000*960
=1152mm2
Area of Steel at Support (Bottom Bars)
IS 456-2000 𝑓𝑐𝑘 4.6∗𝑀𝑠
Annex G Ast= 0.5* 𝑓𝑦 *( 1-√1 − 𝑓𝑐𝑘∗𝑏∗𝑑2 ) b*d
(G-1.1b) 1548.037 mm2
Providing 16mm Dia rebar
Area of each rebar=201.062
𝑅𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
No. of rebar required= 𝑎𝑟𝑒𝑎 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑟𝑒𝑏𝑎𝑟
1548.037
= 201.062
=7.699 no.
Area of reinforcement provided,
Ast provided=10*201.062
=2010.62mm2
Pt=.2094%
Providing 16mm
Spacing of bars, rebar @100 c/c
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑏𝑎𝑟
Sv= Ast provided * 1000
=100mm
Provide 16mm rebar @100 c/c
IS 456-2000 Area of Steel at mid span (Top rebar)
Annex G 𝑓𝑐𝑘 4.6∗,Msp
Ast= 0.5* *( 1-√1 − ) b*d
𝑓𝑦 𝑓𝑐𝑘∗𝑏∗𝑑2
=1283.645mm2
𝑅𝑒𝑖𝑛𝑓𝑜𝑟𝑐𝑒𝑚𝑒𝑛𝑡 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
No. of rebar required=
𝑎𝑟𝑒𝑎 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑟𝑒𝑏𝑎𝑟
1283.645
=
201.062
=6.384
Area of reinforcement provided,
Ast provided=10*201.062
=2010.62mm2
Pt=.2094%
Spacing of bars,
𝐴𝑟𝑒𝑎 𝑜𝑓 𝑟𝑒𝑏𝑎𝑟
Sv= Ast provided * 1000
=100mm Providing 16mm
Provide 16mm rebar @100 c/c rebar @100 c/c
8 Check for one way shear
IS 456-2000 .6*q*l=536.65
Table 19 .4*q*l=357.7678
Shear at critical section
Vu=354.96KN
a/2+d=1.1m
536.65
354.96
1.1 357.77
3.249
22.8238
Vc=378KN
Pt=.4188% Hence,Safe
tc=.42
then,
tc*b*d=.42*1000*900=378KN
